STAT3801_2011Unit4b

STAT3801_2011Unit4b - THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801 ADVANCED LIFE CONTINGENCIES Unit 4b 2010

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801 ADVANCED LIFE CONTINGENCIES Unit 4b 2010- 11 2 nd semester 4.14 Valuation of benefits in multiple decrement context Discrete life insurance for decrement j APV = ∞ X k =0 b k +1 · v k +1 · k | q ( j ) x Continuous life insurance for decrement j APV = Z ∞ b t e- δt · t p ( τ ) x μ ( j ) x ( t ) dt Life annuity-due ¨ a ( τ ) x = ∞ X k =0 1 · v k k p ( τ ) x Actuarial present values and numerical evaluation A = r X j =1 Z ∞ B ( j ) x + t v t t p ( τ ) x μ ( j ) x ( t ) dt Double indemnity provision for n- year term insurance A = 2 Z n v t t p ( τ ) x μ (1) x ( t ) dt + Z n v t t p ( τ ) x μ (2) x ( t ) dt S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 2 Numerical evaluation assuming UDD in multiple decrement model 2 Z n v t t p ( τ ) x μ (1) x ( t ) dt = 2 n- 1 X k =0 v k k p ( τ ) x Z 1 v s s p ( τ ) x + k μ (1) x ( k + s ) ds = 2 n- 1 X k =0 v k +1 k p ( τ ) x q (1) x + k Z 1 (1 + i ) 1- s ds = 2 i δ n- 1 X k =0 v k +1 k p ( τ ) x q (1) x + k Applying similar argument for second integral and combining A = i δ ( n- 1 X k =0 v k +1 k p ( τ ) x h 2 q (1) x + k + q (2) x + k i ) = i δ n- 1 X k =0 v k +1 k p ( τ ) x h q (1) x + k + q ( τ ) x + k i = A (1) 1 x : n + A 1 x : n S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 3 If accidental death benefit = t and no other death benefits A = Z ∞ t v t t p ( τ ) x μ (1) x ( t ) dt = ∞ X k =0 v k k p ( τ ) x Z 1 ( k + s ) v s s p ( τ ) x + k μ (1) x ( k + s ) ds = ∞ X k =0 v k +1 k p ( τ ) x q (1) x + k Z 1 ( k + s )(1 + i ) 1- s ds = ∞ X k =0 v k +1 k p ( τ ) x q (1) x + k i δ k + 1 δ- 1 i ¶ Using midpoint rule for approximate integration Z 1 ( k + s )(1 + i ) 1- s ds ≈ k + 1 2 ¶ v- 1 2 A = ∞ X k =0 v k + 1 2 k p ( τ ) x q (1) x + k k + 1 2 ¶ For varying benefits assuming UDD of j th decrement in multiple decrement model ∞ X k =0 v k +1 k p ( τ ) x q ( j ) x + k Z 1 B ( j ) x + k + s (1 + i ) 1- s ds = ∞ X k =0 v k + 1 2 k p ( τ ) x q ( j ) x + k B ( j ) x + k + 1 2 by midpoint integration rule. S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 4 Example 4.9: In single decrement model for decrement 3, 1 2 at midyear and 1 2 at year end, UDD for other single decrements 1 2 q (1) x + k 1 2 q (1) x + k 1 2 q (2) x + k 1 2 q (3) x + k 1 2 q (2) x + k 1 2 q (3) x + k ← - - - - - - -- → ↓ ← - - - - - - -- → ↓ | | | | | time 1 4 1 2 3 4 1 age x + k x + k + 1 4 x + k + 1 2 x + k + 3 4 x + k + 1 Actuarial present value A = ∞ X k =0 v k k p ( τ ) x h ( 1 2 + ) q (3) x + k v 1 2 B (3) x + k + 1 2 + v 1 2 ( 1 2 + ) p ( τ ) x + k · ( 1 2 + ) q (3) x + k + ( 1 2 + ) v 1 2 B (3) x + k +1 ‚ Now ( 1 2 + ) q (3) x + k = ( 1 2- ) p ( τ ) x + k • 1...
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This note was uploaded on 07/30/2011 for the course STAT 3801 taught by Professor Kc during the Fall '11 term at HKU.

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STAT3801_2011Unit4b - THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801 ADVANCED LIFE CONTINGENCIES Unit 4b 2010

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